Question

A fighter plane is pulling out for drive at $900km/h$ in a vertical circle of radius $2km$. Its mass is $5000kg$. Find the force exerted by the air on it at the lowest point.

• A. $2.0625\times {10}^{4}N$ upward
• B. $2.0625\times {10}^{5}N$ downward
• C. $2.0625\times {10}^{5}N$ upward
• D. $2.0625\times {10}^{4}N$ downward

Answer 1

Answer: (C)

$2.0625\times {10}^{5}N$ upward

We know, $v=900km/h=250m/s$

So, force exerted by the air at the lowest point:

$\frac{{mv}^2}{r}+mg=\frac{5\times {10}^3\times {\left(250\right)}^2}{2\times {10}^3}+5\times {10}^4$ = $2.0625$ $\times {10}^{5}N$ upward.